Vertical motion with air friction

[cscott]

An object falling from some point in the air (near the surface) with air friction, R = kv.

So,

$$m\frac{dv}{dt} = mg - kv$$

Seperate variables for

$$\int \frac{m}{mg - kv} \cdot dv = \int dt$$

and for the LHS I use integration by parts, so,

$$(m)(mgv - \frac{1}{2}kv^2) - \int mgv - \frac{1}{2}kv^2 \cdot dv = t$$

am on the right track here?

 

[nrqed]

An object falling from some point in the air (near the surface) with air friction, R = kv.

So,

$$m\frac{dv}{dt} = mg - kv$$

Seperate variables for

$$\int \frac{m}{mg - kv} \cdot dv = \int dt$$

and for the LHS I use integration by parts, so,

$$(m)(mgv - \frac{1}{2}kv^2) - \int mgv - \frac{1}{2}kv^2 \cdot dv = t$$

am on the right track here?

Your integration by parts is incorrect (I have no idea what you used for ''u'' and ''v'' but there is no way to get what you wrote).
The integral over v is basically a log, that's all.

 

[cscott]

Woops, I see what I did. Let me try that again.

But it should take the form

$$(m)\int (mg - kv)^{-1} \cdot dv - \int (1)\left(\int (mg - kv)^{-1}\right) \cdot dv = t$$

correct?

edit: missed your part about a log. I guess we agree?

 

[nrqed]

Woops, I see what I did. Let me try that again.

But it should take the form

$$(m)\int (mg - kv)^{-1} \cdot dv - \int (1)\left(\int (mg - kv)^{-1}\right) \cdot dv = t$$

correct?

edit: missed your part about a log. I guess we agree?

? The first term is the integral you started with! I am not sure where the second term could come from.

You can do the integral in one step, without using integ by parts

 

[cscott]

asdfakj I wasn't thinking of that m as a constant for some reason :\ I see now, thanks.